Show that the equality $$\cos\dfrac{\gamma}{2}=\sqrt{\dfrac{p(p-c)}{ab}}$$ holds for a $\triangle ABC$ with sides $AB=c,BC=a, AC=b$, semi-perimeter $p$ and $\measuredangle ACB=\gamma$.
I have just proved that $$l_c=\dfrac{2ab\cos\dfrac{\gamma}{2}}{a+b}$$
Is this a well known formula for angle bisectors?
Can we use it somehow?
Thank you in advance!
On
Your formula for length of angle bisectors is correct. To derive for sine/cosine of half-angles, a straightforward way is using the formulas : $$1+\cos 2\theta = 2\cos^2 \theta \, , \quad 1-\cos 2\theta = 2\sin^2 \theta$$
By cosine-rule in $\triangle ABC$, $$\cos \gamma = \frac{a^2+b^2-c^2}{2ab}$$
$$\Rightarrow \cos \frac{\gamma}{2}=\sqrt{\frac{1+\cos \gamma}{2}}$$ which simplifies to
$$ \cos \frac{\gamma}{2}=\sqrt{\frac{p(p-c)}{ab}}$$
Similarly you can also find $\sin (\gamma/2)$, $\tan (\gamma/2)$ etc.
On
We have standard formula for $R = {abc\over 4S}$ and $S = rp$, where $S$ is area and $p$ perimeter of triangle.
If we multiply these two we get $$\cos ^2 {\gamma\over 2} = {S(p-c)\over rab} = {p(p-c)\over ab}$$ (Remember that $\sin \gamma = 2\sin{\gamma\over 2}\cos {\gamma \over 2}$)
On
As for the last part (how we can use the bisector).. The property that the ratio of sides is the ratio in which opposite side $c$ is intersected
$$ e=\dfrac{AC}{ CB}=\dfrac{ AL}{ LB},$$
a constant that defines the Apollonius circle of constant $e$. Bisection can be done either for the internal or for external angle.
Form half-angle trig-identity
$$\cos\gamma=2\cos^2\frac{\gamma}{2}-1\iff \cos \frac{\gamma}{2}=\sqrt{\frac{1+\cos\gamma}{2}}$$
Substituting the value from cosine formula: $\cos \gamma=\frac{a^2+b^2-c^2}{2ab}$ $$\cos \frac{\gamma}{2}=\sqrt{\frac{1+\frac{a^2+b^2-c^2}{2ab}}{2}}$$ $$=\sqrt{\frac{(a+b)^2-c^2}{4ab}}$$
$$=\sqrt{\frac{(a+b+c)(a+b-c)}{4ab}}$$ $$=\sqrt{\frac{\left(\frac{a+b+c}{2}\right)\left(\frac{a+b+c}{2}-c\right)}{ab}}$$ $$=\sqrt{\frac{p(p-c)}{ab}}$$